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Question
The density of iron crystal is 8.54-gram cm–3. If the edge length of the unit cell is 2.8 A° and atomic mass is 56 gram mol–1, find the number of atoms in the unit cell. (Given: Avogadro's number = 6.022x1023,1A° = 1x10-8 cm)
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Solution
Given a = 2.8Å = `2.8 xx 10^(-8)` cm
Molar Mass = 56 gm
Density = 8.54 g cm-3
Volume = (a)3
`= (2.8 xx 10^(-8))^3`
`= 21.95 xx 10^(-24) cm^3`
Density of unit cell = `"Mass of unit cell"/"Volume of unit cell"`
∴ Mass of unit cell = density x volume
`= 21.95 xx 10^(-24) xx 8.54`
`= 187.47 xx 10^(-24)`
56 gm of Fe contains 6.022 x 1023 atoms
`187.47 xx 10^(-24)` gm Fe contain = `(6.022xx10^23 xx 187.47xx10^(-24))/56`
= `20.15xx 10^(-1)`
= 2.01
= 2
Number of atoms in unit cell is 2.
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